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Bernoulli Polynomials, Fourier Series and Zeta Numbers

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Bernoulli Polynomials, Fourier Series and Zeta Numbers International Journal of Pure and Applied Mathematics Volume 88 No. 1 2013, 65-75 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu AP doi: http://dx.doi.org/10.12732/ijpam.v88i1.5 ijpam.eu BERNOULLI POLYNOMIALS, FOURIER SERIES AND ZETA NUMBERS Ernst E. Scheufens DTU Compute Technical University of Denmark DK-2800 Kgs. Lyngby, DENMARK Abstract: Fourier series for Bernoulli polynomials are used to obtain infor- mation about values of the Riemann zeta function for integer arguments greater than one. If the argument is even we recover the well-known exact values, if the argument is odd we find integral representations and rapidly convergent series. AMS Subject Classification: 33, 40, 41, 42 Key Words: Bernoulli polynomials, Fourier series, zeta numbers, integral representations, rapidly convergent series 1. Bernoulli Polynomials and Fourier Series The Bernoulli polynomials Bn(x) are for x ∈ R defined by the generating func- tion ∞ text tn = Bn(x) , |t| < 2π. (1) et − 1 n! n=0 X The polynomials have the form n n − B (x) = B xn j, x ∈ R, n ∈ N , (2) n j j 0 j X=0 where B0,B1,B2, ... are the Bernoulli numbers defined by (1) with x = 0, i.e. c 2013 Academic Publications, Ltd. Received: June 18, 2013 url: www.acadpubl.eu 66 E.E. Scheufens Bn = Bn(0). The first Bernoulli polynomial is B0(x) = 1, and the next three are as follows, 1 B (x) = x − 1 ,B (x) = x2 − x + ,B (x) = x3 − 3 x2 + 1 x. 1 2 2 6 3 2 2 The periodic Bernoulli functions B˜n(x) has period 1, and is for n ∈ N defined by B˜n(x) = Bn(x), 0 ≤ x < 1, B˜n(x + 1) = B˜n(x), x ∈ R. (3) The functions B˜n(x) are continous with continous derivatives up to order n − 1 if n ≥ 2. Fourier series for the periodic Bernoulli functions are for even index ∞ 2(2n)! cos (2mπx) B˜ (x) = (−1)n+1 , x ∈ R, n ∈ N, (4) 2n 2n m2n (2π) m=1 X and for odd index ∞ n+1 2(2n + 1)! sin(2mπx) B˜ n (x) = (−1) , x ∈ R, n ∈ N. (5) 2 +1 2n+1 m2n+1 (2π) m=1 X This Fourier series can be found in for example [5] or [7], where they are defined for B2n(x) and B2n+1(x) restricted to x ∈ [0, 1]. The aim of this paper is to apply the Fourier series in (5) for finding new in- tegral representations and rapidly convergent series for values of the Riemann zeta function, when the argument is an odd integer greater than one, specially if the argument is 3. In [3] are some examples where Fourier series is applied to find values of the Riemann zeta function, when the argument is an even inte- ger. In the next section the general formula for the values of the Riemann zeta function, when the argument is an even integer, is obtained from the Fourier series (4). In [8] Fourier series is applied to generate some special integrals, sums of series, and combinatorial identities. 2. Zeta Numbers The Riemann zeta function is for ℜs > 1 defined by ∞ 1 ζ(s) = , (6) ms m=1 X BERNOULLI POLYNOMIALS... 67 and the zeta numbers is values of ζ(s) when the argument s ∈ N r {1}, see [5]. Setting x = 0 in (4) we get the well-known formula (2π)2n ζ(2n) = (−1)n+1 B , n ∈ N, (7) 2(2n)! 2n where B2,B4,B6, ... are Bernoulli numbers, and we have used that B˜2n(0) = B2n(0) = B2n. Setting x to a particular value in (5) will not give us any information about ζ(2n + 1), n ∈ N, but using the well-known formula ∞ sin(ux) π dx = , u > 0, (8) x 2 Z0 and the uniform convergens of the series for B˜2n+1(x)/x obtained from (5), we get for n ∈ N: ∞ ∞ B˜ (x) 2(2n + 1)! 1 π 2n+1 dx = (−1)n+1 · x 2n+1 m2n+1 2 0 (2π) m=1 Z X (2n + 1)! = (−1)n+1 ζ(2n + 1). 2(2π)2n Now we have the following information of zeta numbers with odd argument ∞ 2(2π)2n B˜ (x) ζ(2n + 1) = (−1)n+1 2n+1 dx , n ∈ N. (9) (2n + 1)! x Z0 Using that B˜2n+1(x + 1) = B˜2n+1(x) and B˜2n+1(−x) = −B˜2n+1(x) then for n ∈ N we get ∞ ∞ B˜ (x) 1 B (x) 2k+1 B˜ (x) 2n+1 dx = 2n+1 dx + 2n+1 dx x x − x Z0 Z0 k=1 Z2k 1 X∞ 1 B (x) 1 B˜ (x) = 2n+1 dx + 2n+1 dx, x − x + 2k 0 k 1 Z X=1 Z where 1 B˜ (x) 0 B˜ (x) 1 B˜ (x) 2n+1 dx = 2n+1 dx + 2n+1 dx − x + 2k − x + 2k x + 2k Z 1 Z 1 Z0 1 B (x) 1 B (x) = 2n+1 dx + 2n+1 dx. x − 2k x + 2k Z0 Z0 68 E.E. Scheufens Consequently we have ∞ ˜ 1 ∞ 1 B2n+1(x) B2n+1(x) 2xB2n+1(x) N dx = dx + 2 2 dx , n ∈ . (10) 0 x 0 x 0 x − 4k Z Z Xk=1 Z From (9) and (10) we can go two ways, one way to get an integral representation for ζ(2n + 1), and another way to get a series representation for ζ(2n + 1). 3. Integral representations of Zeta(2n+1) Changing the order of summation and integration in (10), and using that ∞ 2x π πx 1 = cot( ) − , x ∈ R r {0, ±2, ±4, · · ·} , x2 − 4k2 2 2 x Xk=1 we get for n ∈ N ∞ 1 1 ∞ B˜2n+1(x) B2n+1(x) 2xB2n+1(x) dx = dx + 2 2 dx, 0 x 0 x 0 x − 4k ! Z Z Z Xk=1 where 1 ∞ 1 ∞ 2xB2n+1(x) 2x dx = B n (x) dx x2 − 4k2 2 +1 x2 − 4k2 0 k ! 0 k ! Z X=1 Z X=1 1 π πx 1 = B (x) cot( ) − dx. 2n+1 2 2 x Z0 Consequently we have ∞ B˜ (x) π 1 πx 2n+1 dx = B (x) cot( )dx , n ∈ N. (11) x 2 2n+1 2 Z0 Z0 Inserting (11) in (9) we get the integralrepresentation (2π)2n+1 1 πx ζ(2n + 1) = (−1)n+1 B (x) cot( )dx, n ∈ N. (12) 2(2n + 1)! 2n+1 2 Z0 1 In [2] we have for δ = 1 or δ = 2 the intgral representations 2n+1 δ n+1 (2π) ζ(2n + 1) = (−1) B n (x) cot(πx)dx, n ∈ N. (13) 2δ(2n + 1)! 2 +1 Z0 BERNOULLI POLYNOMIALS... 69 Because πx sin(πx) cos(πx) 1 cot( ) − cot(πx) = − = , x∈ / Z, 2 1 − cos(πx) sin(πx) sin(πx) we get for n ∈ N : 1 1 1 πx B2n+1(x) B n (x) cot( )dx − B n (x) cot(πx)dx = dx. (14) 2 +1 2 2 +1 sin(πx) Z0 Z0 Z0 Inserting the Fourier series (5) we get 1 B (x) 2n+1 dx = 0, n ∈ N, (15) sin(πx) Z0 because 1 sin(2mπx) dx = 0 , m ∈ N. sin(πx) Z0 From(14) and (15) it follows that (12) is equal to (13) if δ = 1. Integration of B˜2n+1(x)/x can be done in an alternative way than we have seen in the previous. This will lead us to an integral representation which involve the Gamma function. The alternative way is ∞ ∞ B˜ (x) k+1 B˜ (x) 2n+1 dx = 2n+1 dx x x Z0 k=0 Zk X∞ 1 B (x) = 2n+1 dx. 0 x + k Xk=0 Z Further we have ∞ 1 ∞ 1 B2n+1(x) 1 1 dx = − − B n (x)dx = x + k k + 1 x + k 2 +1 k=0 Z0 k=0 Z0 X ∞ X 1 1 1 − − B n (x)dx k + 1 x + k 2 +1 Z0 k=0 X ∞ 1 1 1 − −γ + − B (x)dx, k + 1 x + k 2n+1 0 ( k ) Z X=0 70 E.E. Scheufens where we have changed the order of summation and integration in the second line, and the Euler constant γ is subtracted from the sum in the third line. Two times we have used that 1 B2n+1(x)dx = 0. Z0 The digamma function is defined by ∞ Γ′(z) 1 1 ψ(z) = = −γ + − , z ∈ C r {0, −1, −2, ...}, Γ(z) k + 1 z + k Xk=0 and therefore ∞ B˜ (x) 1 2n+1 dx = − ψ(x)B (x)dx, n ∈ N. (16) x 2n+1 Z0 Z0 Using partial integration on the integral in (16) we get ∞ B˜ (x) 1 2n+1 dx = (2n + 1) B (x) ln Γ(x)dx, n ∈ N, (17) x 2n Z0 Z0 ′ where we have used that B2n+1(x) = (2n + 1)B2n(x), and lim (ln Γ(x)B2n+1(x)) = lim (ln Γ(x)B2n+1(x)) = 0. t→1− t→0+ Inserting (17) in (9) we get 2(2π)2n 1 ζ(2n + 1) = (−1)n+1 B (x) ln Γ(x)dx , n ∈ N. (18) (2n)! 2n Z0 Compared with (12) and (13) we see that involving the gamma function, we get B2n(x) instead of B2n+1(x) in the integrand. 4. Series Representations of Zeta(2n+1) From (2) we can find formulas for the integrals on the righthand side of (10).
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